Find the sum to infinity of a geometric series having a second term of -9 and fifth term of 1/3 g

S infinity= -51/4

Step-by-step explanation:

First term

ar= -9

Fifth term

ar⁴= 1/3

Solving for the value of the r and a

ar= -9

ar⁴= 1/3

Dividing each other

r³=1/3 * -1/9

r³=-1/27

r= 3√-1/27

r= -1/3

Solving for a

ar= -9

a(-1/3)= 9

a= 9*-3

a= -27

Sum of a go to infinity is given by the formula

S infinity= a/(1-r)

S infinity= -27/(1-(-1/3))

S infinity= -27/(1+1/3)

S infinity= -27 * 3/4

S infinity= -51/4

step-by-step explanation:

the options are kind of confusing. i can tell you what i see going on here when i look at the graph.

the class and exam medians are closely related, and it looks like the low outlier on the exam did pull the median lower. [tex][/tex]

mxb67

step-by-step explanation: